problem 7

Internal problem ID [5685]
Internal ﬁle name [OUTPUT/4933_Sunday_June_05_2022_03_10_57_PM_13379888/index.tex]

Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number: 7.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeﬀ"

\[ \boxed {y^{\prime \prime }+7 y^{\prime }+12 y=21 \,{\mathrm e}^{3 t}} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {7}{2}}, y^{\prime }\left (0\right ) = -10\right ] \end {align*}

6.7.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end {align*}

Where here \begin {align*} p(t) &=7\\ q(t) &=12\\ F &=21 \,{\mathrm e}^{3 t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+7 y^{\prime }+12 y = 21 \,{\mathrm e}^{3 t} \end {align*}

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (y\right ) =Y(s) \end {align*}

Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right ) &= s Y(s) - y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+7 s Y \left (s \right )-7 y \left (0\right )+12 Y \left (s \right ) = \frac {21}{s -3}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&={\frac {7}{2}}\\ y'(0) &=-10 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-\frac {29}{2}-\frac {7 s}{2}+7 s Y \left (s \right )+12 Y \left (s \right ) = \frac {21}{s -3} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {7 s^{2}+8 s -45}{2 \left (s -3\right ) \left (s^{2}+7 s +12\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{2 s -6}+\frac {1}{2 s +6}+\frac {5}{2 \left (s +4\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{2 s -6}\right ) &= \frac {{\mathrm e}^{3 t}}{2}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s +6}\right ) &= \frac {{\mathrm e}^{-3 t}}{2}\\ \mathcal {L}^{-1}\left (\frac {5}{2 \left (s +4\right )}\right ) &= \frac {5 \,{\mathrm e}^{-4 t}}{2} \end {align*}

Adding the above results and simplifying gives \[ y=\cosh \left (3 t \right )+\frac {5 \,{\mathrm e}^{-4 t}}{2} \] Simplifying the solution gives \[ y = \cosh \left (3 t \right )+\frac {5 \,{\mathrm e}^{-4 t}}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \cosh \left (3 t \right )+\frac {5 \,{\mathrm e}^{-4 t}}{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \cosh \left (3 t \right )+\frac {5 \,{\mathrm e}^{-4 t}}{2} \] Veriﬁed OK.

6.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+7 y^{\prime }+12 y=21 \,{\mathrm e}^{3 t}, y \left (0\right )=\frac {7}{2}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-10\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+7 r +12=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +4\right ) \left (r +3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4, -3\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-3 t}+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=21 \,{\mathrm e}^{3 t}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{-4 t} & {\mathrm e}^{-3 t} \\ -4 \,{\mathrm e}^{-4 t} & -3 \,{\mathrm e}^{-3 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{-7 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-21 \,{\mathrm e}^{-4 t} \left (\int {\mathrm e}^{7 t}d t \right )+21 \,{\mathrm e}^{-3 t} \left (\int {\mathrm e}^{6 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{3 t}}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-3 t}+\frac {{\mathrm e}^{3 t}}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-3 t}+\frac {{\mathrm e}^{3 t}}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {7}{2} \\ {} & {} & \frac {7}{2}=c_{1} +c_{2} +\frac {1}{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-4 c_{1} {\mathrm e}^{-4 t}-3 c_{2} {\mathrm e}^{-3 t}+\frac {3 \,{\mathrm e}^{3 t}}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-10 \\ {} & {} & -10=-4 c_{1} -3 c_{2} +\frac {3}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {5}{2}, c_{2} =\frac {1}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{7 t}+{\mathrm e}^{t}+5\right ) {\mathrm e}^{-4 t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{7 t}+{\mathrm e}^{t}+5\right ) {\mathrm e}^{-4 t}}{2} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`

\[ y \left (t \right ) = \frac {5 \,{\mathrm e}^{-4 t}}{2}+\cosh \left (3 t \right ) \]

6.7 problem 7

6.7.1 Existence and uniqueness analysis

6.7.2 Maple step by step solution